NCERT Solutions for Class 11 Maths Chapter 12: Limits and Derivatives

These Class 11 Maths Chapter 12 solutions cover Limits and Derivatives — your first taste of Calculus. Every question of Exercise 12.1, Exercise 12.2 and the Miscellaneous Exercise is reproduced verbatim from the NCERT textbook (Reprint 2026–27) and solved step by step, with the algebra of limits, trigonometric limits, derivatives from first principle and the product & quotient rules worked out in full.

Class: 11 Subject: Mathematics Chapter: 12 – Limits and Derivatives Exercises: 12.1, 12.2 & Miscellaneous Session: 2026–27
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Chapter 12 Overview

Chapter 12, Limits and Derivatives, is the gateway to Calculus — the branch of mathematics that studies how the value of a function changes as the input changes. The chapter begins with an intuitive idea of the derivative through average velocity, then gives a naive definition of a limit, the left- and right-hand limits, and the algebra of limits. It evaluates limits of polynomials, rational functions and trigonometric functions (using the famous results sin x/x → 1 and (1 − cos x)/x → 0), and the standard limit (xn − an)/(x − a). It then defines the derivative as a limit (the first principle), develops the algebra of derivatives (sum, difference, product and quotient rules) and obtains derivatives of standard functions such as xn, sin x and tan x.

Key Concepts & Definitions

Limit of a function: as x → a, if f(x) → l then l is the limit, written limx→a f(x) = l. It is the value f(x) is expected to take near a, whether or not f(a) is defined.

Left-hand limit (LHL): the value f(x) approaches as x → a from values less than a, written limx→a− f(x).

Right-hand limit (RHL): the value f(x) approaches as x → a from values greater than a, written limx→a+ f(x). The limit exists only when LHL = RHL.

Algebra of limits: the limit of a sum, difference, product or quotient (denominator ≠ 0) of two functions equals the corresponding operation on their limits.

Sandwich (Squeeze) Theorem: if f(x) ≤ g(x) ≤ h(x) near a and lim f = lim h = l, then lim g = l.

Derivative at a point: f′(a) = limh→0 [f(a + h) − f(a)] / h, provided the limit exists. Geometrically it is the slope of the tangent at (a, f(a)).

Derivative function (first principle): f′(x) = limh→0 [f(x + h) − f(x)] / h, also written d/dx (f(x)) or dy/dx.

Important Formulas (Chapter 12)

Standard limit: limx→a (xn − an)/(x − a) = n·an−1 (true for any real n, a > 0).

Trigonometric limits: limx→0 (sin x)/x = 1  •  limx→0 (1 − cos x)/x = 0  •  limx→0 (tan x)/x = 1.

Polynomial/rational limit: limx→a f(x) = f(a) when f is a polynomial; for a rational function take f(a) if the denominator ≠ 0, else cancel common factors first.

First principle: f′(x) = limh→0 [f(x + h) − f(x)] / h.

Power rule: d/dx (xn) = n·xn−1 (any real n).

Sum/difference: (u ± v)′ = u′ ± v′.   Product (Leibnitz): (uv)′ = u′v + uv′.   Quotient: (u/v)′ = (u′v − uv′)/v2.

Standard derivatives: d/dx(sin x) = cos x • d/dx(cos x) = −sin x • d/dx(tan x) = sec2x • d/dx(constant) = 0.

Exercise 12.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

Evaluate the following limits in Exercises 1 to 22.

1. limx→3 (x + 3)

SOLUTION Substitute x = 3 directly (polynomial): 3 + 3 = 6.

2. limx→π (x − 22/7)

SOLUTION Substitute x = π: π − 22/7 = π − 22/7.

3. limr→1 πr2

SOLUTION Substitute r = 1: π(1)2 = π.

4. limx→4 (4x + 3)/(x − 2)

SOLUTION Denominator at x = 4 is 2 ≠ 0, so substitute: (4·4 + 3)/(4 − 2) = 19/2. ∴ the limit is 19/2.

5. limx→−1 (x10 + x5 + 1)/(x − 1)

SOLUTION Denominator at x = −1 is −2 ≠ 0, so substitute: ((−1)10 + (−1)5 + 1)/(−1 − 1) = (1 − 1 + 1)/(−2) = 1/(−2). ∴ the limit is −1/2.

6. limx→0 [(x + 1)5 − 1]/x

SOLUTION Put y = x + 1, so y → 1 as x → 0, and x = y − 1. The limit becomes limy→1 (y5 − 15)/(y − 1). Using (xn − an)/(x − a) → n·an−1: = 5·(1)4 = 5.

7. limx→2 (3x2 − x − 10)/(x2 − 4)

SOLUTION At x = 2 it is 0/0. Factorise: 3x2 − x − 10 = (x − 2)(3x + 5); x2 − 4 = (x − 2)(x + 2). Cancel (x − 2): limx→2 (3x + 5)/(x + 2) = (6 + 5)/(2 + 2) = 11/4. ∴ the limit is 11/4.

8. limx→3 (x4 − 81)/(2x2 − 5x − 3)

SOLUTION At x = 3 it is 0/0. Numerator x4 − 81 = (x − 3)(x + 3)(x2 + 9); denominator 2x2 − 5x − 3 = (x − 3)(2x + 1). Cancel (x − 3): limx→3 (x + 3)(x2 + 9)/(2x + 1) = (6)(18)/(7) = 108/7. ∴ the limit is 108/7.

9. limx→0 (ax + b)/(cx + 1)

SOLUTION Denominator at x = 0 is 1 ≠ 0, so substitute: (a·0 + b)/(c·0 + 1) = b/1. ∴ the limit is b.

10. limz→1 (z1/3 − 1)/(z1/6 − 1)

SOLUTION At z = 1 it is 0/0. Divide numerator and denominator by (z − 1) and use (zn − 1)/(z − 1) → n (with a = 1). = limz→1 [(z1/3 − 1)/(z − 1)] ÷ [(z1/6 − 1)/(z − 1)] = (1/3) ÷ (1/6) = (1/3)·6 = 2. ∴ the limit is 2.

11. limx→1 (ax2 + bx + c)/(cx2 + bx + a),   a + b + c ≠ 0

SOLUTION Substitute x = 1: numerator = a + b + c, denominator = c + b + a = a + b + c. Since a + b + c ≠ 0, the ratio is (a + b + c)/(a + b + c) = 1.

12. limx→−2 (1/x + 1/2)/(x + 2)

SOLUTION Combine the numerator: 1/x + 1/2 = (2 + x)/(2x). So the expression = (x + 2)/(2x) ÷ (x + 2) = 1/(2x). As x → −2: 1/(2·(−2)) = −1/4. ∴ the limit is −1/4.

13. limx→0 (sin ax)/(bx)

SOLUTION Write (sin ax)/(bx) = (sin ax)/(ax) · (a/b). As x → 0, ax → 0 so (sin ax)/(ax) → 1. ∴ the limit is 1·(a/b) = a/b.

14. limx→0 (sin ax)/(sin bx),   a, b ≠ 0

SOLUTION = limx→0 [(sin ax)/(ax)·ax] / [(sin bx)/(bx)·bx] = [1·ax]/[1·bx] = a/b. ∴ the limit is a/b.

15. limx→π sin(π − x)/[π(π − x)]

SOLUTION Put y = π − x, so y → 0 as x → π. The limit = limy→0 (sin y)/(πy) = (1/π)·limy→0 (sin y)/y = (1/π)·1. ∴ the limit is 1/π.

16. limx→0 (cos x)/(π − x)

SOLUTION No 0/0 here. Substitute x = 0: cos 0/(π − 0) = 1/π. ∴ the limit is 1/π.

17. limx→0 (cos 2x − 1)/(cos x − 1)

SOLUTION Use cos 2x − 1 = −2 sin2x and cos x − 1 = −2 sin2(x/2). Ratio = sin2x / sin2(x/2) = [(sin x)/x]2·x2 ÷ { [(sin(x/2))/(x/2)]2·(x/2)2 }. As x → 0 the sine ratios → 1, leaving x2/(x/2)2 = x2/(x2/4) = 4. ∴ the limit is 4.

18. limx→0 (ax + x cos x)/(b sin x)

SOLUTION Divide numerator and denominator by x: numerator → a + cos x, denominator = b·(sin x)/x. = limx→0 (a + cos x) / [b·(sin x)/x] = (a + 1)/(b·1) = (a + 1)/b. ∴ the limit is (a + 1)/b.

19. limx→0 x·sec x

SOLUTION x·sec x = x/cos x. As x → 0: 0/cos 0 = 0/1 = 0. ∴ the limit is 0.

20. limx→0 (sin ax + bx)/(ax + sin bx),   a, b, a + b ≠ 0

SOLUTION Divide top and bottom by x: numerator = a·(sin ax)/(ax) + b, denominator = a + b·(sin bx)/(bx). As x → 0 the sine ratios → 1: = (a + b)/(a + b) = 1. ∴ the limit is 1.

21. limx→0 (cosec x − cot x)

SOLUTION cosec x − cot x = (1 − cos x)/sin x = [(1 − cos x)/x] ÷ [(sin x)/x]. As x → 0: (0)/(1) = 0. ∴ the limit is 0.

22. limx→π/2 (tan 2x)/(x − π/2)

SOLUTION Put y = x − π/2, so y → 0 and 2x = 2y + π. Then tan 2x = tan(2y + π) = tan 2y. Limit = limy→0 (tan 2y)/y = limy→0 [(tan 2y)/(2y)]·2 = 1·2 = 2. ∴ the limit is 2.

23. Find limx→0 f(x) and limx→1 f(x), where f(x) = 2x + 3,   x ≤ 0;   f(x) = 3(x + 1),   x > 0.

SOLUTION At x = 0: LHL = limx→0−(2x + 3) = 3; RHL = limx→0+3(x + 1) = 3(0 + 1) = 3. LHL = RHL, so limx→0 f(x) = 3. At x = 1: x = 1 > 0, so f(x) = 3(x + 1); both one-sided limits = 3(1 + 1) = 6, so limx→1 f(x) = 6. ∴ limx→0 f(x) = 3 and limx→1 f(x) = 6.

24. Find limx→1 f(x), where f(x) = x2 − 1,   x ≤ 1;   f(x) = −x2 − 1,   x > 1.

SOLUTION LHL = limx→1−(x2 − 1) = 1 − 1 = 0. RHL = limx→1+(−x2 − 1) = −1 − 1 = −2. Since LHL (0) ≠ RHL (−2), the limit does not exist at x = 1.

25. Evaluate limx→0 f(x), where f(x) = |x|/x,   x ≠ 0;   f(x) = 0,   x = 0.

SOLUTION For x > 0, |x|/x = 1; for x < 0, |x|/x = −1. RHL = limx→0+ 1 = 1; LHL = limx→0− (−1) = −1. Since LHL ≠ RHL, the limit does not exist at x = 0.

26. Find limx→0 f(x), where f(x) = x/|x|,   x ≠ 0;   f(x) = 0,   x = 0.

SOLUTION For x > 0, x/|x| = 1; for x < 0, x/|x| = −1. RHL = 1, LHL = −1. Since they differ, the limit does not exist at x = 0.

27. Find limx→5 f(x), where f(x) = |x| − 5.

SOLUTION Near x = 5, x is positive, so |x| = x and f(x) = x − 5. limx→5 (x − 5) = 5 − 5 = 0. ∴ the limit is 0.

28. Suppose f(x) = a + bx,   x < 1;   f(x) = 4,   x = 1;   f(x) = b − ax,   x > 1, and if limx→1 f(x) = f(1), what are possible values of a and b?

SOLUTION f(1) = 4. For the limit to exist and equal f(1), LHL = RHL = 4. LHL = limx→1−(a + bx) = a + b. RHL = limx→1+(b − ax) = b − a. So a + b = 4 and b − a = 4. Adding: 2b = 8 ⇒ b = 4; subtracting: 2a = 0 ⇒ a = 0. ∴ a = 0, b = 4.

29. Let a1, a2, …, an be fixed real numbers and define a function f(x) = (x − a1)(x − a2) … (x − an). What is limx→a1 f(x)? For some a ≠ a1, a2, …, an, compute limx→a f(x).

SOLUTION f is a polynomial, so its limit at any point equals its value there. limx→a1 f(x) = (a1 − a1)(a1 − a2) … (a1 − an) = 0 (the first factor is 0). For a ≠ any ai: limx→a f(x) = (a − a1)(a − a2) … (a − an).

30. If f(x) = |x| + 1,   x < 0;   f(x) = 0,   x = 0;   f(x) = x − 1,   x > 0. For what value(s) of a does limx→a f(x) exist?

SOLUTION For x < 0, f(x) = x + 1 (since |x| = −x gives −x + 1… note the book writes f(x) = x + 1 for x < 0). For x > 0, f(x) = x − 1. At a = 0: LHL = limx→0−(x + 1) = 1; RHL = limx→0+(x − 1) = −1. These differ, so the limit does not exist at 0. At any a ≠ 0: f is a polynomial on each side and continuous there, so the limit exists. ∴ limx→a f(x) exists for all a ≠ 0.

31. If the function f(x) satisfies limx→1 [f(x) − 2]/(x2 − 1) = π, evaluate limx→1 f(x).

SOLUTION As x → 1, the denominator x2 − 1 → 0. For the given ratio to tend to a finite value π, the numerator must also → 0. So limx→1 [f(x) − 2] = 0, i.e. limx→1 f(x) − 2 = 0. ∴ limx→1 f(x) = 2.

32. If f(x) = mx2 + n,   x < 0;   f(x) = nx + m,   0 ≤ x ≤ 1;   f(x) = nx3 + m,   x > 1. For what integers m and n do both limx→0 f(x) and limx→1 f(x) exist?

SOLUTION At x = 0: LHL = limx→0−(mx2 + n) = n; RHL = limx→0+(nx + m) = m. For existence, n = m. At x = 1: LHL = limx→1−(nx + m) = n + m; RHL = limx→1+(nx3 + m) = n + m. These are always equal, so the limit at 1 exists for any integers m, n. ∴ limx→0 f(x) exists only when m = n; limx→1 f(x) exists for any integral values of m and n.

Exercise 12.2 Solutions

1. Find the derivative of x2 − 2 at x = 10.

SOLUTION f(x) = x2 − 2 ⇒ f′(x) = 2x. At x = 10: f′(10) = 2(10) = 20.

2. Find the derivative of x at x = 1.

SOLUTION f(x) = x ⇒ f′(x) = 1, a constant. At x = 1: f′(1) = 1.

3. Find the derivative of 99x at x = 100.

SOLUTION f(x) = 99x ⇒ f′(x) = 99 (constant for all x). At x = 100: f′(100) = 99.

4. Find the derivative of the following functions from first principle. (i) x3 − 27   (ii) (x − 1)(x − 2)   (iii) 1/x2   (iv) (x + 1)/(x − 1)

SOLUTION (i) f′(x) = limh→0 [((x + h)3 − 27) − (x3 − 27)]/h = limh→0 [(x + h)3 − x3]/h = limh→0 [3x2h + 3xh2 + h3]/h = limh→0 (3x2 + 3xh + h2) = 3x2. (ii) f(x) = x2 − 3x + 2. f′(x) = limh→0 [( (x+h)2 − 3(x+h) + 2 ) − (x2 − 3x + 2)]/h = limh→0 [2xh + h2 − 3h]/h = limh→0 (2x + h − 3) = 2x − 3. (iii) f(x) = 1/x2. f′(x) = limh→0 [1/(x+h)2 − 1/x2]/h = limh→0 [x2 − (x+h)2] / [h·x2(x+h)2] = limh→0 [−(2x + h)] / [x2(x+h)2] = −2x/x4 = −2/x3. (iv) f(x) = (x + 1)/(x − 1). f(x + h) − f(x) = (x+h+1)/(x+h−1) − (x+1)/(x−1). Common denominator: numerator = (x+h+1)(x−1) − (x+1)(x+h−1) = −2h. So [f(x+h)−f(x)]/h = −2 / [(x+h−1)(x−1)]. As h → 0: −2/(x − 1)2.

5. For the function f(x) = x100/100 + x99/99 + … + x2/2 + x + 1. Prove that f′(1) = 100 f′(0).

SOLUTION Differentiate term by term: d/dx (xk/k) = xk−1. So f′(x) = x99 + x98 + … + x + 1 (the derivative of the constant 1 is 0, and the x term gives 1). f′(0) = 0 + 0 + … + 0 + 1 = 1 (only the last term survives). f′(1) = 199 + 198 + … + 1 + 1 = 100 ones = 100. ∴ f′(1) = 100 = 100·1 = 100 f′(0). Hence proved.

6. Find the derivative of xn + axn−1 + a2xn−2 + … + an−1x + an for some fixed real number a.

SOLUTION Here a is constant, so ak are constants. Differentiate term by term using d/dx(xm) = m·xm−1 (the last term an gives 0). f′(x) = nxn−1 + a(n − 1)xn−2 + a2(n − 2)xn−3 + … + an−1.

7. For some constants a and b, find the derivative of (i) (x − a)(x − b)   (ii) (ax2 + b)2   (iii) (x − a)/(x − b)

SOLUTION (i) Expand: x2 − (a + b)x + ab. Derivative = 2x − (a + b) = 2x − a − b. (ii) Expand: (ax2 + b)2 = a2x4 + 2abx2 + b2. Derivative = 4a2x3 + 4abx = 4ax(ax2 + b). (iii) Quotient rule with u = x − a (u′ = 1), v = x − b (v′ = 1): (u′v − uv′)/v2 = [(x − b) − (x − a)]/(x − b)2 = (a − b)/(x − b)2.

8. Find the derivative of (xn − an)/(x − a) for some constant a.

SOLUTION Quotient rule with u = xn − an (u′ = nxn−1), v = x − a (v′ = 1). f′(x) = [nxn−1(x − a) − (xn − an)·1] / (x − a)2. = [nxn − anxn−1 − xn + an] / (x − a)2.

9. Find the derivative of (i) 2x − 3/4   (ii) (5x3 + 3x − 1)(x − 1)   (iii) x−3(5 + 3x) (iv) x5(3 − 6x−9)   (v) x−4(3 − 4x−5)   (vi) 2/(x + 1) − x2/(3x − 1)

SOLUTION (i) d/dx(2x − 3/4) = 2. (ii) Expand: (5x3 + 3x − 1)(x − 1) = 5x4 − 5x3 + 3x2 − 3x − x + 1 = 5x4 − 5x3 + 3x2 − 4x + 1. Derivative = 20x3 − 15x2 + 6x − 4. (iii) x−3(5 + 3x) = 5x−3 + 3x−2. Derivative = −15x−4 − 6x−3 = −3(5 + 2x)/x4. (iv) x5(3 − 6x−9) = 3x5 − 6x−4. Derivative = 15x4 + 24x−5 = 15x4 + 24/x5. (v) x−4(3 − 4x−5) = 3x−4 − 4x−9. Derivative = −12x−5 + 36x−10 = −12/x5 + 36/x10. (vi) First term 2/(x + 1): derivative = −2/(x + 1)2. Second term x2/(3x − 1): by quotient rule = [2x(3x − 1) − x2(3)]/(3x − 1)2 = (3x2 − 2x)/(3x − 1)2 = x(3x − 2)/(3x − 1)2. So the answer is −2/(x + 1)2 − x(3x − 2)/(3x − 1)2.

10. Find the derivative of cos x from first principle.

SOLUTION f′(x) = limh→0 [cos(x + h) − cos x]/h. Use cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2). = limh→0 [−2 sin(x + h/2) sin(h/2)]/h = limh→0 [−sin(x + h/2)]·[sin(h/2)/(h/2)]. As h → 0, sin(h/2)/(h/2) → 1 and sin(x + h/2) → sin x: f′(x) = −sin x.

11. Find the derivative of the following functions: (i) sin x cos x   (ii) sec x   (iii) 5 sec x + 4 cos x   (iv) cosec x (v) 3 cot x + 5 cosec x   (vi) 5 sin x − 6 cos x + 7   (vii) 2 tan x − 7 sec x

SOLUTION (i) sin x cos x = (1/2)sin 2x, or by product rule: cos x·cos x + sin x·(−sin x) = cos2x − sin2x = cos 2x. (ii) sec x = 1/cos x; quotient rule gives sin x/cos2x = sec x tan x. (iii) 5(sec x tan x) + 4(−sin x) = 5 sec x tan x − 4 sin x. (iv) cosec x = 1/sin x; quotient rule gives −cos x/sin2x = −cosec x cot x. (v) 3(−cosec2x) + 5(−cosec x cot x) = −3 cosec2x − 5 cosec x cot x. (vi) 5 cos x − 6(−sin x) + 0 = 5 cos x + 6 sin x. (vii) 2(sec2x) − 7(sec x tan x) = 2 sec2x − 7 sec x tan x.

Miscellaneous Exercise on Chapter 12 — Solutions

1. Find the derivative of the following functions from first principle: (i) −x   (ii) (−x)−1   (iii) sin(x + 1)   (iv) cos(x − π/8)

SOLUTION (i) f′(x) = limh→0 [−(x + h) − (−x)]/h = limh→0 (−h)/h = −1. (ii) f(x) = −1/x. f′(x) = limh→0 [−1/(x + h) + 1/x]/h = limh→0 [x − (x + h)] / [−h·x(x + h)]·(−1)… computing: = limh→0 (1/h)·[−1/(x+h) + 1/x] = limh→0 (1/h)·[h/(x(x+h))] = 1/x2. ∴ 1/x2. (iii) f′(x) = limh→0 [sin(x + 1 + h) − sin(x + 1)]/h = limh→0 [2 cos(x + 1 + h/2) sin(h/2)]/h = cos(x + 1)·1 = cos(x + 1). (iv) Similarly, with the argument shifted by −π/8: f′(x) = limh→0 [cos(x − π/8 + h) − cos(x − π/8)]/h = −sin(x − π/8).

Find the derivative of the following functions (a, b, c, d, p, q, r, s are fixed non-zero constants; m and n are integers).

2. (x + a)

SOLUTION d/dx (x + a) = 1.

3. (px + q)(r/x + s)

SOLUTION Expand: (px + q)(r/x + s) = pr + psx + qr/x + qs. Derivative = ps + qr·(−1/x2) = −qr/x2 + ps.

4. (ax + b)(cx + d)2

SOLUTION Product rule with u = ax + b (u′ = a) and v = (cx + d)2 (v′ = 2c(cx + d)). = a(cx + d)2 + (ax + b)·2c(cx + d) = 2c(ax + b)(cx + d) + a(cx + d)2.

5. (ax + b)/(cx + d)

SOLUTION Quotient rule: [a(cx + d) − (ax + b)c]/(cx + d)2 = (ad − bc)/(cx + d)2. ∴ (ad − bc)/(cx + d)2.

6. (1 + 1/x)/(1 − 1/x)

SOLUTION Multiply top and bottom by x: = (x + 1)/(x − 1), x ≠ 0, 1. Quotient rule: [(x − 1) − (x + 1)]/(x − 1)2 = −2/(x − 1)2. ∴ −2/(x − 1)2, x ≠ 0, 1.

7. 1/(ax2 + bx + c)

SOLUTION Quotient rule with u = 1, v = ax2 + bx + c (v′ = 2ax + b): = [0·v − 1·(2ax + b)]/v2. ∴ −(2ax + b)/(ax2 + bx + c)2.

8. (ax + b)/(px2 + qx + r)

SOLUTION Quotient rule: [a(px2 + qx + r) − (ax + b)(2px + q)]/(px2 + qx + r)2. Numerator = apx2 + aqx + ar − (2apx2 + aqx + 2bpx + bq) = −apx2 − 2bpx + ar − bq. ∴ (−apx2 − 2bpx + ar − bq)/(px2 + qx + r)2.

9. (px2 + qx + r)/(ax + b)

SOLUTION Quotient rule: [(2px + q)(ax + b) − (px2 + qx + r)a]/(ax + b)2. Numerator = 2apx2 + 2bpx + aqx + bq − apx2 − aqx − ar = apx2 + 2bpx + bq − ar. ∴ (apx2 + 2bpx + bq − ar)/(ax + b)2.

10. a/x4 − b/x2 + cos x

SOLUTION Write as ax−4 − bx−2 + cos x. Derivative = a(−4)x−5 − b(−2)x−3 − sin x. ∴ −4a/x5 + 2b/x3 − sin x.

11. 4√x − 2

SOLUTION f(x) = 4x1/2 − 2. Derivative = 4·(1/2)x−1/2 = 2x−1/2 = 2/√x.

12. (ax + b)n

SOLUTION Using the chain rule (or expanding by binomial theorem and differentiating): d/dx (ax + b)n = n(ax + b)n−1·a. ∴ na(ax + b)n−1.

13. (ax + b)n(cx + d)m

SOLUTION Product rule with u = (ax + b)n (u′ = na(ax + b)n−1), v = (cx + d)m (v′ = mc(cx + d)m−1). = na(ax + b)n−1(cx + d)m + mc(ax + b)n(cx + d)m−1. Factor: (ax + b)n−1(cx + d)m−1[mc(ax + b) + na(cx + d)].

14. sin(x + a)

SOLUTION Expand sin(x + a) = sin x cos a + cos x sin a (cos a, sin a constants). Derivative = cos x cos a − sin x sin a = cos(x + a).

15. cosec x cot x

SOLUTION Product rule with u = cosec x (u′ = −cosec x cot x), v = cot x (v′ = −cosec2x). = (−cosec x cot x)cot x + cosec x(−cosec2x) = −cosec x cot2x − cosec3x. ∴ −cosec3x − cosec x cot2x.

16. cos x/(1 + sin x)

SOLUTION Quotient rule: [(−sin x)(1 + sin x) − cos x(cos x)]/(1 + sin x)2. Numerator = −sin x − sin2x − cos2x = −sin x − 1 = −(1 + sin x). = −(1 + sin x)/(1 + sin x)2 = −1/(1 + sin x).

17. (sin x + cos x)/(sin x − cos x)

SOLUTION Quotient rule: u = sin x + cos x (u′ = cos x − sin x), v = sin x − cos x (v′ = cos x + sin x). Numerator = (cos x − sin x)(sin x − cos x) − (sin x + cos x)(cos x + sin x) = −(sin x − cos x)2 − (sin x + cos x)2. = −[(1 − sin 2x) + (1 + sin 2x)] = −2. ∴ −2/(sin x − cos x)2.

18. (sec x − 1)/(sec x + 1)

SOLUTION Quotient rule with u = sec x − 1, v = sec x + 1, both having derivative sec x tan x. Numerator = sec x tan x(sec x + 1) − (sec x − 1)sec x tan x = sec x tan x[(sec x + 1) − (sec x − 1)] = 2 sec x tan x. ∴ 2 sec x tan x/(sec x + 1)2.

19. sinnx

SOLUTION By the chain rule, d/dx (sin x)n = n(sin x)n−1·cos x. ∴ n sinn−1x cos x.

20. (a + b sin x)/(c + d cos x)

SOLUTION Quotient rule: u = a + b sin x (u′ = b cos x), v = c + d cos x (v′ = −d sin x). Numerator = b cos x(c + d cos x) − (a + b sin x)(−d sin x) = bc cos x + bd cos2x + ad sin x + bd sin2x = bc cos x + ad sin x + bd. ∴ (bc cos x + ad sin x + bd)/(c + d cos x)2.

21. sin(x + a)/cos x

SOLUTION Quotient rule: u = sin(x + a) (u′ = cos(x + a)), v = cos x (v′ = −sin x). Numerator = cos(x + a)cos x − sin(x + a)(−sin x) = cos(x + a)cos x + sin(x + a)sin x = cos((x + a) − x) = cos a. ∴ cos a/cos2x.

22. x4(5 sin x − 3 cos x)

SOLUTION Product rule: u = x4 (u′ = 4x3), v = 5 sin x − 3 cos x (v′ = 5 cos x + 3 sin x). = 4x3(5 sin x − 3 cos x) + x4(5 cos x + 3 sin x). = x3[(20 sin x − 12 cos x) + (5x cos x + 3x sin x)] = x3(5x cos x + 3x sin x + 20 sin x − 12 cos x).

23. (x2 + 1)cos x

SOLUTION Product rule: u = x2 + 1 (u′ = 2x), v = cos x (v′ = −sin x). = 2x cos x + (x2 + 1)(−sin x) = −x2 sin x − sin x + 2x cos x.

24. (ax2 + sin x)(p + q cos x)

SOLUTION Product rule: u = ax2 + sin x (u′ = 2ax + cos x), v = p + q cos x (v′ = −q sin x). = (2ax + cos x)(p + q cos x) + (ax2 + sin x)(−q sin x). ∴ −q sin x(ax2 + sin x) + (p + q cos x)(2ax + cos x).

25. (x + cos x)(x − tan x)

SOLUTION Product rule: u = x + cos x (u′ = 1 − sin x), v = x − tan x (v′ = 1 − sec2x = −tan2x). = (1 − sin x)(x − tan x) + (x + cos x)(−tan2x). ∴ −tan2x(x + cos x) + (x − tan x)(1 − sin x).

26. (4x + 5 sin x)/(3x + 7 cos x)

SOLUTION Quotient rule: u = 4x + 5 sin x (u′ = 4 + 5 cos x), v = 3x + 7 cos x (v′ = 3 − 7 sin x). Numerator = (4 + 5 cos x)(3x + 7 cos x) − (4x + 5 sin x)(3 − 7 sin x). Expanding and simplifying gives 35 + 15x cos x + 28 cos x + 28x sin x − 15 sin x. ∴ (35 + 15x cos x + 28 cos x + 28x sin x − 15 sin x)/(3x + 7 cos x)2.

27. x2cos(π/4)/sin x

SOLUTION cos(π/4) is a constant; pull it out: cos(π/4)·d/dx (x2/sin x). Quotient rule on x2/sin x: [2x sin x − x2cos x]/sin2x = x(2 sin x − x cos x)/sin2x. ∴ x cos(π/4)(2 sin x − x cos x)/sin2x.

28. x/(1 + tan x)

SOLUTION Quotient rule: u = x (u′ = 1), v = 1 + tan x (v′ = sec2x). = [1·(1 + tan x) − x·sec2x]/(1 + tan x)2. ∴ (1 + tan x − x sec2x)/(1 + tan x)2.

29. (x + sec x)(x − tan x)

SOLUTION Product rule: u = x + sec x (u′ = 1 + sec x tan x), v = x − tan x (v′ = 1 − sec2x). = (x + sec x)(1 − sec2x) + (x − tan x)(1 + sec x tan x). ∴ (x + sec x)(1 − sec2x) + (x − tan x)(1 + sec x tan x).

30. x/sinnx

SOLUTION Quotient rule: u = x (u′ = 1), v = sinnx (v′ = n sinn−1x cos x). = [1·sinnx − x·n sinn−1x cos x] / sin2nx = [sin x − nx cos x]·sinn−1x / sin2nx. ∴ (sin x − nx cos x)/sinn+1x.

Common Mistakes to Avoid

Watch out for these

  • Concluding a limit exists without checking that LHL = RHL — for piecewise functions and modulus functions, always compute both one-sided limits.
  • Substituting directly when the result is 0/0 — first factorise (or use the standard limit (xn − an)/(x − a)) and cancel the offending factor.
  • Forgetting to convert to the standard form before using (sin x)/x → 1 — the angle inside sin must match the denominator (multiply/divide by the right factor).
  • Mixing up the quotient rule order: it is (u′v − uv′)/v2, not (uv′ − u′v)/v2.
  • Sign errors with trig derivatives: d/dx(cos x) = −sin x, d/dx(cosec x) = −cosec x cot x, d/dx(cot x) = −cosec2x.
  • In first-principle problems, dropping the limit symbol too early — keep limh→0 until after the h in the denominator has cancelled.

Practice MCQs & Assertion–Reason

1. limx→0 (sin x)/x equals:

(a) 0    (b) 1    (c) ∞    (d) does not exist

2. limx→2 (x2 − 4)/(x − 2) equals:

(a) 0    (b) 2    (c) 4    (d) does not exist

3. The value of limx→a (xn − an)/(x − a) is:

(a) n·an    (b) n·an−1    (c) an−1    (d) n·an+1

4. d/dx (tan x) equals:

(a) sec x tan x    (b) −cosec2x    (c) sec2x    (d) cot x

5. limx→0 (1 − cos x)/x equals:

(a) 1    (b) 0    (c) 1/2    (d) −1

6. The derivative of the constant function f(x) = 7 is:

(a) 7    (b) 1    (c) 0    (d) x

7. If f(x) = |x|/x for x ≠ 0, then limx→0 f(x):

(a) is 1    (b) is −1    (c) is 0    (d) does not exist

8. The quotient rule states (u/v)′ equals:

(a) u′/v′    (b) (u′v − uv′)/v2    (c) (uv′ − u′v)/v2    (d) u′v′

9. The derivative of cos x is:

(a) sin x    (b) −sin x    (c) −cos x    (d) sec2x

10. limx→0 (tan x)/x equals:

(a) 0    (b) 1    (c) −1    (d) ∞

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(b), 6-(c), 7-(d), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: limx→0 |x|/x does not exist.

Reason: The left-hand limit (−1) and the right-hand limit (+1) are unequal.

A-R 2. Assertion: For f(x) = x2, f′(x) = 2x.

Reason: The derivative of xn is n·xn−1 for any positive integer n.

A-R 3. Assertion: limx→1 (x10 + x5 + 1)/(x − 1) can be found by direct substitution.

Reason: The denominator x − 1 is zero at x = 1.

A-R 4. Assertion: d/dx (sin x cos x) = cos 2x.

Reason: sin x cos x = (1/2)sin 2x and the derivative of the product follows from the product rule.

A-R 5. Assertion: The limit of a function at a point and the value of the function at that point are always equal.

Reason: A limit describes the value a function approaches near a point, which may differ from its actual value there.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(D).

Quick Revision Summary

  • The limit exists only when LHL = RHL; the limit and the function value at a point may differ.
  • Algebra of limits: limits respect +, −, × and ÷ (denominator ≠ 0).
  • For polynomials limx→a f(x) = f(a); for rational functions cancel common factors when the form is 0/0.
  • Standard limits: (xn − an)/(x − a) → n·an−1, (sin x)/x → 1, (1 − cos x)/x → 0.
  • Derivative (first principle): f′(x) = limh→0 [f(x + h) − f(x)]/h; it is the slope of the tangent.
  • Rules: (u ± v)′ = u′ ± v′; (uv)′ = u′v + uv′; (u/v)′ = (u′v − uv′)/v2.
  • Standard derivatives: (xn)′ = nxn−1, (sin x)′ = cos x, (cos x)′ = −sin x, (tan x)′ = sec2x.

How to score full marks in this chapter

For limit questions, always substitute first; only if you hit 0/0 should you factorise, rationalise or convert to a standard trig limit. Memorise the three standard limits and the basic derivatives, and quote the rule (product/quotient/first principle) you are using so each step earns its mark. In first-principle problems write the full quotient, simplify the numerator, cancel h, then take h → 0. Keep your algebra tidy — most marks here are lost to sign slips, not to ideas.

Frequently Asked Questions

What is Class 11 Maths Chapter 12 about?

Chapter 12, Limits and Derivatives, is an introduction to Calculus. It covers the intuitive idea of a derivative, the definition and algebra of limits, left- and right-hand limits, limits of polynomial, rational and trigonometric functions, and the definition of the derivative (first principle) together with the sum, difference, product and quotient rules and derivatives of standard functions.

How many exercises are there in Class 11 Maths Chapter 12?

There are two main exercises — Exercise 12.1 (limits) and Exercise 12.2 (derivatives) — plus a Miscellaneous Exercise on Chapter 12. Every question of all three is solved step by step on this page.

What are the most important limits to remember in this chapter?

The three standard limits are limx→a (xn − an)/(x − a) = n·an−1, limx→0 (sin x)/x = 1 and limx→0 (1 − cos x)/x = 0. Most exercise problems reduce to one of these.

Are these Class 11 Maths Chapter 12 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with answers verified against the book’s answer key.

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